16.4. INTERSECTIONS OF Na(L) WITH CONJUGATES OF Ca(L) 1183In this section we develop some technical tools, which we apply in the final
section to show that J 1 is the only group satisfying Hypothesis 16.1.1.
Set K := Cc(L), H := Nc(L), and H* := H/K. As Lis standard in G,
H = Nc(K). Thus for each nontrivial 2-subgroup X of K, Nc(X)::::; H by Remark
16.3.3. In particular,
Gz::SH.
For K' E Kc, define L(K') := £9, where g E G with K9 = K'; as Nc(K) =Nc(L) = H, this definition is independent of the choice of g, and the set of such
elements is a coset of H in G.
Recall TL =TnL and Tc =TnK.
Our discussion in this section will be based on an analysis of the set
~ = ~(K) := {K' EKG - {K}: INK1(K)l2 > 1}.
To see that ~ is nonempty under our hypotheses, we appeal to I.8.2: Since G is
simple, K is not normal in G = (KG). Therefore if ~ is empty, then H is strongly
embedded in G by I.8.2. Then by the Bender-Suzuki classification (see Theorem
SE on p. 20 of [GLS99]) of simple groups with strongly embedded subgroups,
G = 021 (G) is a Bender group, contrary to our assumption in Hypothesis 16.1.1
that G is not of even characteristic. Thus we conclude that
~ is nonempty.Recall that in Notation 16.2.2, T E Syb(G), and then T E Syb(H) using
Theorem 16.2.4. Then as ~ is nonempty we can extend that earlier Notation by
adopting:
NOTATION 16.4.1. K' E ~' L' := L(K'), H' := NG(K'), andR E Syl2(NK1(K))
with R::::; T. For each involution r in R, set Lr := 02 (CL(r)). Also set H* := H/ K.
Since R::::; Tin Notation 16.4.1, R normalizes TL and Tc by Theorem 16.2.4.
Our next result lists elementary properties of the members of ~(K):
LEMMA 16.4.2. {1) R ~ Nr 0 (R) = Cr 0 (R) = Nr 0 (K') E Syb(NK(K')), with
Rn K = 1. In particular R is faithful on L and INK^1 (K)l2 = INK(K')l2·
(2) L 1, H'.
{3)R=K'nT.
(4) There exists g E G with K' = K^9 and Nr(K')::::; T^9.(5) For each 1 =/= X::::; R, NG(X)::::; H'.
{6) If Nr(R) E Syb(NH(R)), then Cr(R(z)) E Syl2(CG(R(z))).
PROOF. Part (5) is a restatement of Remark 16.3.3. We apply parts (1) and
(2) of I.7.7 with K', K, TcR in the roles of "K, K^9 , S" to obtain Nr 0 (R) =
Cr 0 (R) ~ R. By I.7.7.3, Nr 0 (R) is Sylow in NK(K'), completing the proof of (1).
By (1), INK(K')I is even, so (2) follows from 16.3.13.3. Part (3) holds since
R E Syb(NK1(K)) and R::::; T. Let g E G with K' = K^9 and Nr(K') ::S T' E
Syb(H'). As T9 E Syb(H') there is y EH' with T9Y = T', so replacing g by gy,
we may take Nr(K')::::; T9. Thus (4) holds.
If Nr(R) E Syl 2 (NH(R)) then Cr(R) E Syb(CH(R)), so as z E Z(T),